Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

EDIT: I meant to have the coefficients reversed, showing: $$\frac{n}{n-1}(1-(1-x)^n)^n + (1-x)^{n-1} \leq 1$$ This version should be true.. but still trying to prove it...

ORIGINAL: Is it possible to show: $$(1-(1-x)^n)^n + \frac{n}{n-1}(1-x)^{n-1} \leq 1$$ for $0<x<1$ and $n\geq 2$ (and $n$ is an integer)? This increases the difficulty over the other questions I just asked.

the second term seems to decrease with $n$, so I can solve for the minimum of $n$. But the first term doesn't completely increase with $n$ -- the lines cross when I plot the first term with $n=2$ and then $n=3$. So I'm not sure where to go from there.

share|improve this question
1  
If $0 < x < 1$ then you can simply replace $y = 1 - x$ and say $0 < y < 1$, making the inequality a bit easier. You could then maybe also replace $z = y^n$ such that again $0 < z < 1$, but I'm not sure that helps. –  TMM Sep 14 '11 at 23:11
1  
The $n/(n-1)$ reminds me of Hölder's inequality. –  Srivatsan Sep 14 '11 at 23:20
    
Is $n \geq 2$ or $n > 2$? The title and the (original) question text say different things. –  Srivatsan Sep 15 '11 at 13:07
    
@Srivatsan: Honestly, I would be happy with either! –  Angada Sep 15 '11 at 16:34
add comment

2 Answers

up vote 3 down vote accepted

It fails already for $n=2$. Make the substitution suggested by Thijs, and the left-hand side becomes $(1-y^2)^2+2y = 1+y^4 +2y(1-y) > 1$ for $0<y<1$.

share|improve this answer
    
Thank you!! Turns out, I accidentally wrote the wrong equation, but the comments helped me figure it out anyway. –  Angada Sep 15 '11 at 1:23
add comment

This might be false for every positive integer $n \ge 2$.

I believe we can show this for $\displaystyle x \gt 1 - \frac{1}{n-1}$ by using Bernoulli's inequality on the first term on the left side.

Using Bernoulli's.

$$(1 - (1-x)^n)^n \ge 1 - n(1-x)^n$$

And so

$$(1 - (1-x)^n)^n + \frac{n(1-x)^{n-1}}{n-1} \ge 1 - n(1-x)^n + \frac{n(1-x)^{n-1}}{n-1}$$

$$= 1 - n(1-x)^{n-1}\left((1-x) - \frac{1}{n-1}\right) $$

if $$ 1 - x \lt \frac{1}{n-1}$$

then

$$1 - n(1-x)^{n-1}\left((1-x) - \frac{1}{n-1}\right) \gt 1$$

Are you hoping for this to be true, or do you know this to be true (like assigned textbook problem).

share|improve this answer
    
Thank you!! Turns out, I accidentally wrote the wrong equation, so this one was false, but the comments helped me figure it out how to prove the one I meant to write, which I know to be true. –  Angada Sep 15 '11 at 1:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.